Question: The graphs of $2y + x + 3 = 0$  and $3y + ax + 2 = 0$ are perpendicular. Solve for $a.$
Explanation: Solving $2y + x + 3 = 0$ for $y$ gives $ y = \frac{-1}{2}x - \frac{3}{2},$ so the slope of this line is $-\frac{1}{2}.$

Solving $3y + ax + 2 = 0$ for $y$ gives $ y = \frac{-a}{3}x - \frac{2}{3},$ so the slope of this line is $- \frac{a}{3}.$

In order for these lines to be perpendicular, we must have $$\left(-\frac{1}{2}\right)\left(-\frac{a}{3}\right) = -1.$$Solving for $a$ gives $a = \boxed{-6}.$